CDF of a normal vector is there a closed form?

[dabeth]

Suppose a vector X of length n, where each component of X is normally distributed with mean 0 and variance 1, and independent of the other components. I want to know the probability that at least one of X_1>2, X_2>2.5, X_3>1.9, etc. happens (inclusive), i.e. the probability that the vector's individual values are all greater than a vector with some arbitrary values. In the above case, I'd be looking for the probability that the individual components of X are greater than [2, 2.5, 1.9, ... ]. To me this looks like a CDF, but a CDF of n independently distributed random variables. Does this have a closed form? If not, is there a manageable way to approximate it? Would it have a sigmoid shape (i.e. concave where all components are positive in n-space)?

Actually, I think the vector notation confuses things. So basically, I'm just asking about the "CDF" of a bunch of independent random variables that all are distributed standard normal.

Thanks.

 

[bpet]

If you put B_1 = 1 if X_1>2 or 0 otherwise, etc, you have a collection of Bernoulli random variables with parameters p_1=Prob[X_1>2] etc, so the probability that at least one is 1 is 1-(1-p_1)*...*(1-p_n).

 

[chiro]

Suppose a vector X of length n, where each component of X is normally distributed with mean 0 and variance 1, and independent of the other components. I want to know the probability that at least one of X_1>2, X_2>2.5, X_3>1.9, etc. happens (inclusive), i.e. the probability that the vector's individual values are all greater than a vector with some arbitrary values. In the above case, I'd be looking for the probability that the individual components of X are greater than [2, 2.5, 1.9, ... ]. To me this looks like a CDF, but a CDF of n independently distributed random variables. Does this have a closed form? If not, is there a manageable way to approximate it? Would it have a sigmoid shape (i.e. concave where all components are positive in n-space)?

Actually, I think the vector notation confuses things. So basically, I'm just asking about the "CDF" of a bunch of independent random variables that all are distributed standard normal.

Thanks.

Hello dabeth and welcome to the forums.

In terms of length, I'll assume three dimensions in which L^2 = (X1)^2 + (X2)^2 + (X3)^2 and L^2 = N^2 for some given N.

In terms of the random variable L^2, assuming X1, X2, X3 are distributed N(0,1), then L^2 has a Chi-Square distribution with 3 degrees of freedom (This is the definition of a chi-square distribution). In this case it is better to deal with the distribution of the square of the length rather than the length itself, because you have a known distribution that you can use. You can use to get a distribution for the length of your vector for some given range of the length.

You can't just get the probability of the length being some fixed real number, because the chi-square distribution is continuous and the probability in this regard is zero. You have to make the interval have some width of some sort (Perhaps some multiple of an integer).

With regards to finding out whether |X1| > 2.5, this is going to be a conditional probability.

So let's say you choose that you are considering values between a and a+1 for length of the vector. What you will need to find out is P(|X1| > 2.5 | a^2 < X1^2 + X2^2 + X3^2 < (a+1)^2) which is going to be P(|X1| > 2.5 AND a^2 < Chi-Square(3) < (a+1)^2) / P(a^2 < Chi-Square(3) < (a+1)^2).

Since these are actual probabilities: you can calculate them to get a specific value for your real probability. (You will have to use either a computer or get some statistical tables that have detailed probabilities that are more extensive than your standard 0.01,0.025,0.05,0.1 etc) [I would use a computer to do it].

Just one note though: in terms of this we are dealing with the squared value of the component so you need to make sure you are calculating the right thing. In my example I used the absolute value |X1| because of the nature of the squaring effect.

In terms of more complicated probability calculations, you will need to use probability axioms to reduce whatever conditional probabilities you have into ones that you can measure.

Hopefully this has given you some hints to work from.

 

[bpet]

Hello dabeth and welcome to the forums.

In terms of length, I'll assume three dimensions in which L^2 = (X1)^2 + (X2)^2 + (X3)^2 and L^2 = N^2 for some given N.

...

I think he meant the MATLAB length, i.e. the number of elements in the vector.

 

[blue_raver22]

I would like to clarify that there is no specific "CDF of a normal vector" as mentioned in the question. A CDF (cumulative distribution function) is a mathematical function that describes the probability of a random variable taking on a certain value or less. It is typically used to describe the distribution of a single random variable, not a vector of multiple random variables.

In this case, the vector X is composed of n independent normal random variables, each with mean 0 and variance 1. The probability of at least one of the components of X being greater than a specific value can be calculated using the complement of the joint probability of all components being less than or equal to that value. This can be expressed as:

P(at least one component > a) = 1 - P(X_1 ≤ a, X_2 ≤ a, ..., X_n ≤ a)

Since the components of X are independent, the joint probability can be calculated by multiplying the individual probabilities, which in this case is the CDF of a standard normal distribution. However, this does not have a closed form solution and would require numerical integration or approximation methods to calculate.

In terms of the shape of this "CDF," it would not have a sigmoid shape as it is not a single variable but a function of multiple variables. The shape would depend on the specific values chosen for a and the number of components in the vector X.

In conclusion, while there is no closed form solution for the probability of at least one component of a vector of independent normal random variables being greater than a certain value, it can be calculated using numerical methods and depends on the specific values chosen.